标签:des com http blog style class div img code java c
题目:
You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output:
7 -> 0 –> 8
解题思路:
这题相当于两个大数相加,只不过这里采用的链表的形式,而不是字符串。
解题时最需注意的是,最后一个节点要考虑会不会进位,over =1时,需要增加一个节点。
实现代码:
#include <iostream> using namespace std; /** You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 */ struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { if(l1 == NULL && l2 == NULL) return NULL; ListNode *l3 = new ListNode(-1); ListNode *tnode = l3; int over = 0; while(l1 && l2) { int sum = l1->val + l2->val + over; ListNode *node = new ListNode(sum % 10); over = sum / 10; tnode->next = node; tnode = tnode->next; l1 = l1->next; l2 = l2->next; } if(l1 == NULL && l2 == NULL && over)//后一个节点,要考虑有没进位 { ListNode *node = new ListNode(over); tnode->next = node; return l3->next; } ListNode *left = l1; if(l2) left = l2; while(left) { int sum = left->val + over; ListNode *node = new ListNode(sum % 10); over = sum / 10; tnode->next = node; tnode = tnode->next; left = left->next; } if(over)//同样,最后一个节点,要考虑有没进位 { ListNode *node = new ListNode(over); tnode->next = node; } return l3->next; } }; int main(void) { return 0; }
LeetCode4:Add Two Numbers,码迷,mamicode.com
标签:des com http blog style class div img code java c
原文地址:http://www.cnblogs.com/mickole/p/3699093.html